LDMOS power amplifier

Overcurrent Protection

Like most of us, I use one of these HP server power supplies, which can deliver 50 volts with 60 amps of power. The 50 volts are still low voltage, but the maximum current of 60A can have a heavy impact. Just imagine a broken cable, a broken-through transistor or similar. The power supply mercilessly pushes this 60A into the load. Only when a short circuit is detected the power supply switches off. If (!) it is detected, because that works only with really low-impedance wiring and a low-impedance error. Otherwise, a continuous current of just under 60A flows and the power supply does not switch off. The consequences are easy to imagine. 60A is enough for welding, as it bangs in the Shack, but heavily.

In order to prevent such danger, there are essentially 2 solutions:

a) The wiring must be made thick
enough that a high short-circuit current can flow in the event of a fault. Only then the fuses in the power supply recognize the fault and switches off.
b) the maximum current should be limited by an additional
hardware to the really required level.

A readily built and tested board is available at www.helitron.de/shop

Overcurrent protection:

The overcurrent protection switch described here is looped into the positive line between the power supply and the power amplifier. With a resistor you can set the required maximum current. If this current is exceeded (even for a short time) then the switch opens permanently. You then have to turn off / on to reset the circuit breaker.


the circuit is based on the high-side switch BTS50085. This type withstands the required voltages and currents.


To switch on, put the line ON on the usual + 12V. I have an "ON" button on the front panel.
This will trip Q2 and pull Pin3 of the BTS50085 to ground. Now the BTS switches on and the supply voltage reaches the PA. At the same time a self-locking is made via R3, so that you can release the ON button again, the BTS remains switched on.
At pin5 the resistor R1 is connected. The BTS50085 generates a current flow through R1, which is derived directly from the load current. It is therefore possible to use the voltage drop across R1 as an indicator of the load current. As soon as the voltage at R1 reaches the threshold voltage of Q3 (about 2 to 2.5V), Q3 becomes conductive, as a result Q2 blocks, causing the BTS50085 to open and the load current
is interrupted.
If the power is interrupted, you have to press the ON button again to turn it on. Of course you should first eliminate the cause of the high current.


The circuit requires matching because both the threshold voltage of Q3 and the current through R1 are subject to greater tolerances. The threshold voltage can be found in the data sheet of the Mosfet used (transfer characteristic).

switch-off current is calculated as follows:
I (off) = U (threshold) * 13000 / R1

where the factor 13000 can be in the range of 11000 to 15000 according to the data sheet of the BTS50085. For my copy
it was 11000.

The best way
is to assemble a 2.2kOhm resistor first and check at which power it is turned off. It will happen somewhere at 6A to maybe 10A. Now you have a clue and can calculate what factor the BTS50085 actually has:

Factor = I (off) * R1 / U (threshold)

Now that you know the factor, you can easily calculate the required resistance with the first formula.
I set the current limit to 30A and had to install a R1 of 820 ohms.